Moment Approximations With a Large Zero-Loss Atom

Why a Zero-Loss Atom Changes the Approximation

An aggregate payment can be zero even when the amount paid on a claim, given that a claim occurs, is strictly positive. For example, a policy may make no payment during a period in which no covered claim occurs. When the chance of no payment is substantial, the aggregate distribution has a point mass at zero. A continuous approximation spreads probability smoothly across a range of values and therefore cannot represent that point mass.

This distinction matters especially for probabilities at zero or near zero. An ordinary normal approximation may assign far less probability to zero than the model does, and a continuous moment-matched distribution assigns exactly zero probability to the single value zero. A useful approach is to preserve the zero-payment probability explicitly and approximate only the distribution conditional on a positive payment.

Definition: Let $S$ be a nonnegative aggregate payment. The zero-loss atom is $p_0=P(S=0)$. If $p_0<1$, the conditional positive-payment distribution is the distribution of $S$ given $S>0$. Its cumulative distribution function is $G(x)=P(S\leq x\mid S>0)$.

For a compound model $S=\sum_{j=1}^{N}Y_j$, assume that the claim payments $Y_j$ are strictly positive whenever a claim occurs. Then $S=0$ if and only if $N=0$, so $p_0=P(N=0)$. If individual payments can themselves be zero, that shortcut need not hold: the zero probability must account for both the frequency and the possibility of zero-valued payments.

What the Atom Does to Moment Approximations

The ordinary normal approximation uses the aggregate mean and variance to approximate the entire distribution by a continuous normal distribution. It can be useful when an aggregate is sufficiently spread out, but it has no atom at zero. The same limitation applies to continuous moment-matched approximations such as gamma or lognormal distributions. Matching moments does not, by itself, preserve the probability of no payment.

For $x\geq 0$, an atom-retaining approximation separates the zero case from positive payments:

P(S\leq x)\ \approx\ p_0+(1-p_0)G(x).

Here $G$ is an approximation to the conditional positive-payment distribution. For $x=0$, the approximation gives $P(S\leq 0)=p_0$, provided the positive-payment component has support strictly above zero. For $x<0$, the probability is zero because $S$ is nonnegative. This mixture structure is useful for probabilities, but it does not turn a fitted continuous component into an exact representation of the original distribution.

Theorem: Write $q=1-p_0=P(S>0)$, and suppose $q>0$. If $m=E[S]$ and $v_S=\operatorname{Var}(S)$, the conditional positive-payment mean and variance are $$ \mu_+=E[S\mid S>0]=\frac{m}{q}, \qquad \sigma_+^2=\operatorname{Var}(S\mid S>0) =\frac{v_S+m^2}{q}-\left(\frac{m}{q}\right)^2. $$

These formulas let us fit a distribution to positive payments using aggregate moments and the known atom probability. For instance, a gamma distribution with shape $\alpha$ and scale $\theta$ can match the conditional mean and variance by setting $\alpha=\mu_+^2/\sigma_+^2$ and $\theta=\sigma_+^2/\mu_+$, when $\sigma_+^2>0$. A gamma component has positive support, so the resulting mixture can retain the zero atom without assigning negative payments to the positive component. This is still an approximation: matching two moments does not ensure that the fitted positive tail or other distributional features match the true ones.

Deriving the Conditional Moments

The conditional moment formulas follow from the fact that the zero case contributes nothing to any positive power of $S$. In particular, $S=S\mathbf{1}_{\{S>0\}}$, where $\mathbf{1}_{\{S>0\}}$ is one when $S>0$ and zero otherwise. Thus $E[S]=qE[S\mid S>0]$, giving the conditional mean.

For the variance, first recover the unconditional second raw moment from $E[S^2]=\operatorname{Var}(S)+(E[S])^2$. Since $S^2=0$ when $S=0$, it follows that $E[S^2]=qE[S^2\mid S>0]$. Divide by $q$ to obtain the conditional second raw moment, then subtract the square of the conditional mean:

E[S^2\mid S>0]=\frac{v_S+m^2}{q}, \qquad \operatorname{Var}(S\mid S>0) =E[S^2\mid S>0]-\mu_+^2.

This calculation requires $q>0$. If $q=0$, the aggregate payment is always zero and there is no conditional positive-payment distribution to fit.

Worked Example: A Bernoulli Claim With Exponential Severity

A hypothetical contract has one possible payment during a period. A payment occurs with probability $0.20$; otherwise, the payment is zero. Given a payment, its amount $Y$ has an exponential distribution with mean $1,000. Find the exact probability that the aggregate payment $S$ is at most $500, and compare it with an ordinary normal approximation. Use the exact model’s zero atom in the first calculation.

Method and assumptions. This is a Bernoulli frequency model with payment probability $0.20$. The conditional severity is exponential with mean $1,000, so its variance is $1{,}000{,}000$ and its second raw moment is $2{,}000{,}000$. There is no interest adjustment because the example concerns payment amounts in one period.

Exact probability retaining the atom. The zero-payment probability is $p_0=0.80$, and the positive-payment probability is $q=0.20$. For an exponential payment with mean $1,000, $P(Y\leq 500)=1-e^{-500/1000}=1-e^{-0.5}=0.3935$, to four decimals. Therefore,

P(S\leq 500)=0.80+0.20(0.3935)=0.8787.

Ordinary normal approximation. The aggregate mean is $m=0.20(1{,}000)=200$. Its second raw moment is $E[S^2]=0.20(2{,}000{,}000)=400{,}000$, so its variance is $400{,}000-200^2=360{,}000$. The standard deviation is $\sqrt{360{,}000}=600$. The normal approximation gives $P(S\leq 500)\approx\Phi((500-200)/600)=\Phi(0.5)=0.6915$, to four decimals.

The ordinary normal estimate is substantially below the exact probability because it does not preserve the 0.80 probability of no payment. The atom-retaining calculation avoids the mistake of treating all outcomes as if they came from one continuous distribution.

Compound Models and the Zero Probability

When individual payments are strictly positive, the frequency model gives $p_0$ directly. For a Poisson frequency with mean $\lambda$, $p_0=e^{-\lambda}$. For a binomial frequency with $n$ trials and claim probability $r$, $p_0=(1-r)^n$. In each case, these formulas rely on the stated frequency model and on every claim producing a positive payment.

The aggregate mean and variance can be used alongside this zero probability. For a compound Poisson model with independent, identically distributed positive severities $Y$, independent of $N\sim\operatorname{Poisson}(\lambda)$, the standard compound moments are $E[S]=\lambda E[Y]$ and $\operatorname{Var}(S)=\lambda E[Y^2]$. Those moments do not determine the full aggregate distribution, but together with $p_0$ they determine the first two conditional positive-payment moments through the formulas above.

Worked Example: A Low-Frequency Compound Poisson Aggregate

A hypothetical portfolio has aggregate payment $S=\sum_{j=1}^{N}400$, where $N\sim\operatorname{Poisson}(0.25)$; every claim payment is exactly $400. Find $P(S=0)$ and $P(S\leq 400)$. Compare the latter with the ordinary normal approximation based on the aggregate mean and variance.

Method and assumptions. Claim frequency is Poisson with parameter $\lambda=0.25$, payments are independent of frequency and are strictly positive, and each payment is $400. Thus $S=400N$ and the zero atom is $P(N=0)$.

The zero-payment probability is $p_0=e^{-0.25}=0.7788$, to four decimals. The event $S\leq 400$ is equivalent to $N\leq 1$. Hence the exact probability is

P(S\leq 400)=P(N=0)+P(N=1) =e^{-0.25}(1+0.25)=0.9735.

For the normal approximation, $E[S]=\lambda E[Y]=0.25(400)=100$. Since $E[Y^2]=400^2=160{,}000$, the aggregate variance is $0.25(160{,}000)=40{,}000$, and the standard deviation is $200$. Therefore, $P(S\leq 400)\approx\Phi((400-100)/200)=\Phi(1.5)=0.9332$, to four decimals.

The exact probability includes both the large zero atom and the probability of exactly one claim. The normal approximation includes neither feature explicitly. At zero itself, the exact probability $P(S\leq 0)$ is $0.7788$, whereas the ordinary normal estimate is $\Phi((0-100)/200)=\Phi(-0.5)=0.3085$. A continuity correction can adjust a boundary for a lattice distribution, but it does not generally replace the need to account for a large zero atom.

Fitting a Positive Component While Retaining the Atom

A practical atom-retaining approximation has two parts: keep $p_0$ from the frequency model or another reliable source, then fit a continuous distribution to $S\mid S>0$. The conditional mean and variance must be calculated from the aggregate moments; fitting directly to the unconditional moments would incorrectly assign those moments to the positive component alone.

For a gamma positive component, the shape-scale parameterization is convenient because its mean is $\alpha\theta$ and its variance is $\alpha\theta^2$. Once $\mu_+$ and $\sigma_+^2$ are known, the matching parameters are $\alpha=\mu_+^2/\sigma_+^2$ and $\theta=\sigma_+^2/\mu_+$. The resulting CDF $G$ is then combined with the atom using $p_0+(1-p_0)G(x)$ for $x\geq0$. A fitted positive component is not guaranteed to reproduce the true distribution, and the approximation is not meaningful if the conditional variance is zero for a continuous gamma fit.

Worked Example: Moment-Matching a Gamma Positive Component

A hypothetical aggregate payment is zero with probability $0.75$. Given a positive payment, it has a gamma distribution with shape $2$ and scale $500. Use the aggregate mean and variance to recover the conditional positive moments, fit a gamma component by moment matching, and estimate $P(S\leq 500)$ with the atom retained. Compare with an ordinary normal approximation.

Method and assumptions. The model is a mixture: $S=0$ with probability $0.75$, and otherwise the positive payment follows the stated gamma distribution. The gamma shape-scale convention gives conditional mean $2(500)=1{,}000$ and conditional variance $2(500^2)=500{,}000$. Thus $p_0=0.75$ and $q=0.25$.

The aggregate mean is $m=q(1{,}000)=250$. The conditional second raw moment is $500{,}000+1{,}000^2=1{,}500{,}000$. Therefore the aggregate second raw moment is $q(1{,}500{,}000)=375{,}000$, and the aggregate variance is $375{,}000-250^2=312{,}500$.

Recover the conditional moments from those aggregate moments:

\mu_+=\frac{250}{0.25}=1{,}000, \qquad \sigma_+^2=\frac{312{,}500+250^2}{0.25}-1{,}000^2 =1{,}500{,}000-1{,}000{,}000=500{,}000.

The matching gamma parameters are $\alpha=1{,}000^2/500{,}000=2$ and $\theta=500{,}000/1{,}000=500$. For a gamma distribution with shape $2$ and scale $500$, the CDF at $500 is $1-e^{-1}(1+1)=0.2642$, to four decimals. Retaining the atom gives $P(S\leq 500)\approx0.75+0.25(0.2642)=0.8161$.

For comparison, the ordinary normal approximation has standard deviation $\sqrt{312{,}500}=559.01699$. Its standardized value at $500 is $(500-250)/559.01699=0.4472$, so the estimated probability is $\Phi(0.4472)=0.6726$, to four decimals. In this example the conditional gamma fit recovers the specified positive component, while the unconditional normal approximation still loses the atom.

Common Mistakes and Practical Checks

Using the unconditional moments for the positive component. The positive component occurs only with probability $q$. First divide the aggregate raw moments by $q$, then calculate conditional variance.
Assuming a continuous fit has an atom at zero. A continuous CDF assigns probability zero to the single point zero. A mixture needs an explicit $p_0$ term.
Applying $P(N=0)$ when zero-valued claims are possible. If claims can produce zero payments, $S=0$ can occur even when $N>0$. Calculate the zero probability from the full model.
Using the mixture formula below zero. A nonnegative aggregate has $P(S\leq x)=0$ for $x<0$; the formula $p_0+qG(x)$ is used for $x\geq0$, with $G(0)=0$ for a strictly positive component.
Assuming a good moment match ensures a good tail estimate. The atom-aware mixture corrects the probability of no payment, but the fitted conditional distribution can still misrepresent skewness or tail behavior.

The central decision is whether the event of interest is sensitive to no payment. For probabilities at zero, or at a small positive threshold when $p_0$ is large, preserve the atom explicitly. For probabilities farther into the positive range, compare the atom-retaining approximation with other suitable methods rather than assuming moment matching alone resolves the approximation error. This approach is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, preserve the zero-payment probability when the requested calculation calls for an atom-retaining approximation. Give probabilities to four decimals.

A hypothetical aggregate payment is zero unless one claim occurs; the claim occurs with probability $0.10$ and pays exactly $300. Find $P(S\leq0)$ and the ordinary normal approximation based on the aggregate mean and variance. (Answer: exact $0.9000$; normal approximation $0.3694$)
Let $N\sim\operatorname{Poisson}(0.40)$, with every claim producing a strictly positive payment. Find the probability of zero aggregate payment. (Answer: $e^{-0.40}=0.6703$)
An aggregate payment is zero with probability $0.60$; conditional on a positive payment, it is exponential with mean $800. Estimate $P(S\leq400)$ while retaining the atom. (Answer: $0.60+0.40(1-e^{-0.5})=0.7574$)
An aggregate has $p_0=0.80$, mean $m=120$, and variance $v_S=93{,}600$. Find the conditional positive-payment mean and variance. (Answer: $\mu_+=600$, $\sigma_+^2=180{,}000$)
An aggregate has zero-payment probability $0.70$, mean $100$, and standard deviation $200$. What is the ordinary normal estimate of $P(S\leq0)$, and why does it fail to retain the atom? (Answer: $\Phi(-0.5)=0.3085$; the continuous normal distribution has no point mass at zero, unlike the model’s $0.7000$ atom.)

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